2x/x-2=5+13x^2/x-2 eching

$$\frac { 2 x } { x - 2 } = 5 + \frac { 13 x ^ { 2 } } { x - 2 }$$

x uchun yechish

$$x=-1$$

$$x=\frac{10}{13}\approx 0,769230769$$

Guruhlash orqali faktorlash qadamlari

x qiymati 2 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini x-2 ga ko'paytirish.

$$2x=\left(x-2\right)\times 5+13x^{2}$$

x-2 ga 5 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2x=5x-10+13x^{2}$$

Ikkala tarafdan 5x ni ayirish.

$$2x-5x=-10+13x^{2}$$

-3x ni olish uchun 2x va -5x ni birlashtirish.

$$-3x=-10+13x^{2}$$

Ikkala tarafdan -10 ni ayirish.

$$-3x-\left(-10\right)=13x^{2}$$

-10 ning teskarisi 10 ga teng.

$$-3x+10=13x^{2}$$

Ikkala tarafdan 13x^{2} ni ayirish.

$$-3x+10-13x^{2}=0$$

Polinomni standart shaklga keltirish uchun uni qayta tartiblang. Shartlarni eng yuqoridan eng pastki qiymat ko'rsatgichiga joylashtirish.

$$-13x^{2}-3x+10=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon -13x^{2}+ax+bx+10 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-3 ab=-13\times 10=-130$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -130-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-130 2,-65 5,-26 10,-13$$

Har bir juftlik yigʻindisini hisoblang.

$$1-130=-129 2-65=-63 5-26=-21 10-13=-3$$

Yechim – -3 yigʻindisini beruvchi juftlik.

$$a=10 b=-13$$

-13x^{2}-3x+10 ni \left(-13x^{2}+10x\right)+\left(-13x+10\right) sifatida qaytadan yozish.

$$\left(-13x^{2}+10x\right)+\left(-13x+10\right)$$

Birinchi guruhda -x ni va ikkinchi guruhda -1 ni faktordan chiqaring.

$$-x\left(13x-10\right)-\left(13x-10\right)$$

Distributiv funktsiyasidan foydalangan holda 13x-10 umumiy terminini chiqaring.

$$\left(13x-10\right)\left(-x-1\right)$$

Tenglamani yechish uchun 13x-10=0 va -x-1=0 ni yeching.

$$x=\frac{10}{13} x=-1$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati 2 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini x-2 ga ko'paytirish.

$$2x=\left(x-2\right)\times 5+13x^{2}$$

x-2 ga 5 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2x=5x-10+13x^{2}$$

Ikkala tarafdan 5x ni ayirish.

$$2x-5x=-10+13x^{2}$$

-3x ni olish uchun 2x va -5x ni birlashtirish.

$$-3x=-10+13x^{2}$$

Ikkala tarafdan -10 ni ayirish.

$$-3x-\left(-10\right)=13x^{2}$$

-10 ning teskarisi 10 ga teng.

$$-3x+10=13x^{2}$$

Ikkala tarafdan 13x^{2} ni ayirish.

$$-3x+10-13x^{2}=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$-13x^{2}-3x+10=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -13 ni a, -3 ni b va 10 ni c bilan almashtiring.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-13\right)\times 10}}{2\left(-13\right)}$$

-3 kvadratini chiqarish.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-13\right)\times 10}}{2\left(-13\right)}$$

-4 ni -13 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9+52\times 10}}{2\left(-13\right)}$$

52 ni 10 marotabaga ko'paytirish.

$$x=\frac{-\left(-3\right)±\sqrt{9+520}}{2\left(-13\right)}$$

9 ni 520 ga qo'shish.

$$x=\frac{-\left(-3\right)±\sqrt{529}}{2\left(-13\right)}$$

529 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-3\right)±23}{2\left(-13\right)}$$

-3 ning teskarisi 3 ga teng.

$$x=\frac{3±23}{2\left(-13\right)}$$

2 ni -13 marotabaga ko'paytirish.

$$x=\frac{3±23}{-26}$$

x=\frac{3±23}{-26} tenglamasini yeching, bunda ± musbat. 3 ni 23 ga qo'shish.

$$x=\frac{26}{-26}$$

26 ni -26 ga bo'lish.

$$x=-1$$

x=\frac{3±23}{-26} tenglamasini yeching, bunda ± manfiy. 3 dan 23 ni ayirish.

$$x=-\frac{20}{-26}$$

\frac{-20}{-26} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{10}{13}$$

Tenglama yechildi.

$$x=-1 x=\frac{10}{13}$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati 2 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini x-2 ga ko'paytirish.

$$2x=\left(x-2\right)\times 5+13x^{2}$$

x-2 ga 5 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$2x=5x-10+13x^{2}$$

Ikkala tarafdan 5x ni ayirish.

$$2x-5x=-10+13x^{2}$$

-3x ni olish uchun 2x va -5x ni birlashtirish.

$$-3x=-10+13x^{2}$$

Ikkala tarafdan 13x^{2} ni ayirish.

$$-3x-13x^{2}=-10$$

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$-13x^{2}-3x=-10$$

Ikki tarafini -13 ga bo‘ling.

$$\frac{-13x^{2}-3x}{-13}=-\frac{10}{-13}$$

-13 ga bo'lish -13 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{3}{-13}\right)x=-\frac{10}{-13}$$

-3 ni -13 ga bo'lish.

$$x^{2}+\frac{3}{13}x=-\frac{10}{-13}$$

-10 ni -13 ga bo'lish.

$$x^{2}+\frac{3}{13}x=\frac{10}{13}$$

\frac{3}{13} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{3}{26} olish uchun. Keyin, \frac{3}{26} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{3}{13}x+\left(\frac{3}{26}\right)^{2}=\frac{10}{13}+\left(\frac{3}{26}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{3}{26} kvadratini chiqarish.

$$x^{2}+\frac{3}{13}x+\frac{9}{676}=\frac{10}{13}+\frac{9}{676}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{10}{13} ni \frac{9}{676} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{3}{13}x+\frac{9}{676}=\frac{529}{676}$$

x^{2}+\frac{3}{13}x+\frac{9}{676} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{3}{26}\right)^{2}=\frac{529}{676}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{3}{26}\right)^{2}}=\sqrt{\frac{529}{676}}$$

Qisqartirish.

$$x+\frac{3}{26}=\frac{23}{26} x+\frac{3}{26}=-\frac{23}{26}$$

Tenglamaning ikkala tarafidan \frac{3}{26} ni ayirish.

$$x=\frac{10}{13} x=-1$$